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Proof step 2 from proof is_array-union-disjoint (
previous step
)
⊦ st1 .refs ∩ st .refs = ∅ ∧ r ∈ st1 ∧ is_array(r, st1) → is_array(r, st1 ∪ st)
Lemma
isarray-union-disjoint